string match string index out of range-CodePudding

Could someone please explain to me why this won't work? I could always copy the answer, but then I wouldn't learn why this didn't work.

a scrrenshot of the problem, my work, and the error

Given 2 strings, a and b, return the number of the positions where they contain the same length 2 substring. So "xxcaazz" and "xxbaaz" yields 3, since the "xx", "aa", and "az" substrings appear in the same place in both strings.

def string_match(a, b):
  
  count = 0
  
  if len(a) or len(b) == 0:
    return count
  
  else:
  
    if len(a)> len(b):
      for n in range(len(a)-2):
        if a[n]==b[n]:
          count  =1
          
    else:
      for n in range(len(b)-2):
        if b[n]==a[n]:
          count =1
          
    return count

Thank you so much!!

I tried using index locations to find matching characters in 2 strings

CodePudding user response：

You should use the length of the shortest string in the loop. So you should switch your condition

from

if len(a) > len(b):
    for n in range(len(a)-1):

if len(a) < len(b):
    for n in range(len(a)-1):

That's for the Index out of range error, yet as mentionned in the comments there are some other bugs in your code

# first condition
if len(a) == 0 or len(b) == 0:
    return 0

Full code

def string_match(a, b):
    count = 0
    
    if len(a) == 0 or len(b) == 0:
        return 0
        
    if len(a) < len(b):
        for n in range(len(a)):
            if a[n] == b[n]:
                count  = 1
    else:
        for n in range(len(b)):
            if a[n] == b[n]:
                count  = 1
    return count

NOTE your code does not answer the given task, take your time to develop a better solution

CodePudding user response：

thank you for your help! I made the following adjustments and the code works!

def string_match(a, b):
    count = 0
    n=0
    
    if len(a) == 0 or len(b) == 0:
        return 0
    else:
      if len(a) < len(b):
          for n in range(len(a)-1):
              if a[n:n 2] == b[n:n 2]:
                  count  = 1
      else:
          for n in range(len(b)-1):
              if a[n:n 2] == b[n:n 2]:
                  count  = 1
      return count